Question

Calculate the volume of hydrogen produced when 27.4g of Barium are added to water .

Answer 1

Solution :

⏭ Given:

✏ 27.4 g of Barium are added to water.

⏭ To Find:

✏ The volume of hydrogen gas produced during the process between barium and water.

⏭ Chemical Reaction:

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{Ba + 2H_2O \: \rightarrow \:Ba(OH)_2 + H_2 }}}}} \: \star$

⏭ Concept:

✏ Mass of one mole of Ba = 137g

✏ Volume of one mole of $H_2$ gas = 22.4L

⏭ Calculation:

$\implies \sf \: 137g \: of \: barium \: gives \: 22.4L \: of \: h2 \: gas \\ \\ \therefore \sf \: 27.4g \: of \: barium \: gives... \\ \\ \rightarrow \sf \: volume \: of \: H_2 \: gas = \frac{27.4 \times 22.4}{137} \\ \\ \rightarrow \sf \: volume \: of \: H_2 \: gas = \frac{613.76}{137} \\ \\ \underline{\boxed{ \bold{ \sf{ \orange{volume \: of \: H_2 \: gas = 4.48 \: L}}}}} \: \red{ \star}$